Solve the simultaneous equations.
$\begin{array}{l}
5x + y = 11 \\
x + y = 3
\end{array}$
The system of linear equations is given by:
\[ \begin{cases} 5x + y = 11 \quad (1) \ x + y = 3 \quad (2) \end{cases} \]
We are looking for the values of \(x\) and \(y\) that satisfy both equations simultaneously. Since both equations are linear and involve two variables, we can use the elimination method to solve for the variables. This involves subtracting one equation from the other to eliminate a variable, allowing us to solve for the remaining one.
To eliminate the variable \(y\), we subtract equation (2) from equation (1):
\[(5x + y) - (x + y) = 11 - 3\]
Simplifying the left side:
\[5x - x + y - y = 8\]
\[4x = 8\]
Dividing both sides by 4, we find:
\[x = 2\]
Now that we have determined \(x = 2\), we can substitute this value back into either of the original equations to solve for \(y\). Using equation (2):
\[x + y = 3\]
Substituting \(x = 2\):
\[2 + y = 3\]
Subtracting 2 from both sides gives:
\[y = 3 - 2 = 1\]
Thus, the solution to the system is \(x = 2\) and \(y = 1\).
Finally, we verify the solution by substituting \(x = 2\) and \(y = 1\) back into equation (1):
\[5(2) + 1 = 10 + 1 = 11\]
The values satisfy both equations in the system. The solution is \(x = 2, y = 1\). The final answer is \(\boxed{x=2, y=1}\).
Systems of linear equations are fundamental in scenarios where multiple variables are constrained by different conditions. For instance, if you are buying supplies for a school project, you might know the total cost of two different bundles of art materials. If Bundle A has 5 markers and 1 pen for \(\$11\), and Bundle B has 1 marker and 1 pen for \(\$3\), you can use this system of equations to determine the individual price of a marker and a pen. Solving this allows you to budget effectively and compare prices across different stores or quantities.